Solve for $x$ : $2x^2 - 40x + 200 = 0$
Dividing both sides by $2$ gives: $ x^2 {-20}x + {100} = 0 $ The coefficient on the $x$ term is $-20$ and the constant term is $100$ , so we need to find two numbers that add up to $-20$ and multiply to $100$ The number $-10$ used twice satisfies both conditions: $ {-10} + {-10} = {-20} $ $ {-10} \times {-10} = {100} $ So $(x - {10})^2 = 0$ $x - 10 = 0$ Thus, $x = 10$ is the solution.